How do I solve the inequation \sqrt{n} < \sqrt{n-1} + .01?

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To solve the inequation \(\sqrt{n} < \sqrt{n-1} + 0.01\), it is suggested to first isolate one of the radicals by rewriting the inequality. Squaring both sides maintains the direction of the inequality, as both sides are positive. After squaring, the equation will still contain a radical, so further manipulation is needed to isolate it again. Squaring both sides a second time will help eliminate the radical and lead to a solvable inequality. The final solution indicates that \(n = 2501\).
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Homework Statement


Somebody can help me with thi following problem:

\sqrt{n} - \sqrt{n-1} < 0,01

Answer: n=2501


Homework Equations





The Attempt at a Solution

 
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We can't help you unless you give us some sort of indication that you've actually tried it.
 
I tried this:

A=n^1/2

B= n-1)^1/2

Then i squared both terms :

[n^1/2 - (n-1)^1/2]² <0,0001

But i couldn't find the answer anyways.Just didn't work this way to me.
 
For one thing, your answer is NOT going to be an equation. Instead, it will be an inequality.

To make things easier on yourself, rewrite your inequality with one of the radicals moved to the other side.
\sqrt{n} &lt; \sqrt{n-1} + .01
Now square both sides. Since you're multiplying each side by a positive number, the direction of the inequality won't change.

When you square both sides, you will end up with another radical. Just move all of the other terms to the other side so that the radical is all by itself on one side, then square both sides again.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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